\(1,\\ a,=2y\left(x+3\right)+z\left(x+3\right)=\left(2y+z\right)\left(x+3\right)\\ b,=\left(4x-x-1\right)\left(4x+x+1\right)=\left(3x-1\right)\left(5x+1\right)\\ c,=\left(x-7\right)\left(x+1\right)\\ d,=x^3-x^2-x^2+x+x-1=\left(x-1\right)\left(x^2-x+1\right)\\ 2,\\ a,\Rightarrow\left(x-1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ b,\Rightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\\ \Rightarrow\left(x+5\right)\left(x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\\ c,\Rightarrow10x^2+9x-10x^2-13x+3=0\\ \Rightarrow-4x=-3\Rightarrow x=\dfrac{3}{4}\)
Bài 1:
a) \(=\left(2xy+6y\right)+\left(3z+xz\right)=2y\left(x+3\right)+z\left(x+3\right)=\left(x+3\right)\left(2y+z\right)\)
b) \(=\left(4x-x-1\right)\left(4x+x+1\right)\)
c) \(=\left(x^2-6x+9\right)-16=\left(x-3\right)^2-16=\left(x-3-4\right)\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)
d) Không thấy đề
Bài 2:
a) \(\Rightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
b) \(\Rightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
c) \(\Rightarrow10x^2+9x-10x^2-13x+3=0\)
\(\Rightarrow-4x=-3\Rightarrow x=\dfrac{3}{4}\)
